Calculate Normal Force

[Calvin Murphy]

I am hoping to settle a debate between two engineers.

I have created a simple diagram of the part of the structure we are investigating.

It can be solved in 5 minutes or less by anyone familiar with static equilibrium problems.

The goal is simple enough: find Fn.

Yet, the discussion over whose solution was correct is still unresolved. I build a rig to physically measure it, but the disagreement persisted. Then I reached out to 5 university professors to see if they would help, but I have not heard a response. So here I am. Please help.

 

[Calvin Murphy]

One quick clarification: Fx is a resulting force, not an input. You do not need to solve for Fx. I drew the vector just to help visualize the reaction force at that surface. Sorry for any confusion.

 

[PMR06]

I'll bite... Fn=10N?

 

[ChorasDen]

Does each member have identical length and stiffness?

 

[EngDM]

I agree that it is 10kN. That is, assuming that isn't a hinge where the load is applied. 5kN/sin(30°) = 10kN

 

[IDS]

I'll bite... Fn=10N?

Yes, assuming we are ignoring non-linear effects (and obviously there isn't enough information to evaluate those), it's hard to see how there could be any dispute that the resultant force has to be 5/sin(30) = 10.

 

[JStephen]

Sum moments about Point A and show that there isn't any tangential force at B, then sum vertical forces to find the 10kN as above.
It seems to me there would be more potential error in going from the actual item to this simplification than in solving the system shown.
It is unstable at the left end, so it requires a weightless skate. If a fly lands on the skate, the system collapses.

 

[Eng16080]

I'm curious what the trick here is (considering you're getting PHDs involved), but I'll take a shot:

• Joint A must be in equilibrium, so at Joint A, the sum of forces in the y direction must be zero.
• If Fny is the vertical component of Fn, then Fny = 5N.
• Due to the geometry, sin30 = Fny / Fn, and thus, Fn = 2 * Fny = 2 * 5N = 10N.

 

[Celt83]

Ha had it wrong, first pass ignored the existence of Fx, reworked and in camp Fn=10n

 

[SWComposites]

if A and B are pinned, isn't this unstable? pt A just snaps thru as the left side slides up.

 

[EngDM]

if A and B are pinned, isn't this unstable? pt A just snaps thru as the left side slides up.

Middle must be rigid connection or a cranked member, otherwise yea it would just fail in the middle and you'd be unstable. But A and B being pins doesn't inherently make it unstable.

 

[IDS]

if A and B are pinned, isn't this unstable? pt A just snaps thru as the left side slides up.

Yes, I should have read the notes, I assumed continuity at point A. If it is pinned at A, it is in static equilibrium in the same way that a free-standing column that is pinned at the base is in equilibrium under a pure axial load; i.e. it is highly unstable.

 

[HTURKAK]

if A and B are pinned, isn't this unstable? pt A just snaps thru as the left side slides up.

If A pinned , yes .. it will be unstable. The OP should clarify that , B is pinned and A is rigid .

 

[human909]

If A pinned , yes .. it will be unstable. The OP should clarify that , B is pinned and A is rigid .

A is pinned as per the post. And in the simple model world in which this post was presented the it doesn't.need to be rigid.

Plug this into your preferred frame tool and you'll likely get the same answer if you don't run a non-linear analysis or a buckling analysis.

 

[HTURKAK]

A is pinned as per the post. And in the simple model world in which this post was presented the it doesn't.need to be rigid.

Plug this into your preferred frame tool and you'll likely get the same answer if you don't run a non-linear analysis or a buckling analysis.

Roller support cannot resist moment at real world . The system is label. The OP should clarify if the roller support is fixed with a guide rail etc for moment resistance.For the case shown at OP's sketch , roller support restrained in X direction only.

 

[Calvin Murphy]

Another clarification I should add: based on the system this represents, the skate roller is in essence constrained to a track so that Fx is always perpendicular to the wall. It can react moments and the system should be stable.

I tried to do this in a short-hand way using a perpendicularity symbol on the member that attaches to the skate. I apologize for the confusion.

 

[Eng16080]

It can react moments and the system should be stable

If the roller can resist moment, then there are a total of 4 unknowns with 3 equations to solve them. Therefore the problem is indeterminate and cannot be solved without additional information (member stiffnesses, lengths, etc.)

I’ll change my answer to: unsolvable

 

[Calvin Murphy]

If the roller can resist moment, then there are a total of 4 unknowns with 3 equations to solve them. Therefore the problem is indeterminate and cannot be solved without additional information (member stiffnesses, lengths, etc.)

I’ll change my answer to: unsolvable

It would only react moments at that interface if there were any, which there are not. Stiffness and lengths are not needed to solve.

 

[Calvin Murphy]

Roller support cannot resist moment at real world . The system is label. The OP should clarify if the roller support is fixed with a guide rail etc for moment resistance.For the case shown at OP's sketch , roller support restrained in X direction only.

View attachment 4289

Thank you for the explanation. I was unaware of this caveat with roller constraints.

 

[Eng16080]

It would only react moments at that interface if there were any, which there are not

Fine, then back to 10N.

I’d question if this statement is really true though. I’ll admit I’m having trouble visualizing this support condition